![Interesting approach proof sum 1/n^4= pi^4/90. No Fourier expansion, no Taylor series. No calculus - YouTube Interesting approach proof sum 1/n^4= pi^4/90. No Fourier expansion, no Taylor series. No calculus - YouTube](https://i.ytimg.com/vi/XRA0OhF7E-g/hqdefault.jpg)
Interesting approach proof sum 1/n^4= pi^4/90. No Fourier expansion, no Taylor series. No calculus - YouTube
![Ankur Handa on Twitter: "@InertialObservr If you plug t = 1/pi in the last two equations you arrive at the result. This one extends from the Basel problem also used to obtain Ankur Handa on Twitter: "@InertialObservr If you plug t = 1/pi in the last two equations you arrive at the result. This one extends from the Basel problem also used to obtain](https://pbs.twimg.com/media/EOmw31NVUAEL9w4.jpg:large)
Ankur Handa on Twitter: "@InertialObservr If you plug t = 1/pi in the last two equations you arrive at the result. This one extends from the Basel problem also used to obtain
What's an intuitive explanation of the following mathematical fact: [math]\displaystyle \sum_{n=1}^{\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}[/math]? - Quora
![Given that sum(n=1)^oo 1/n^2=pi^2/6 and sum(n=1)^oo 1/(n^2+8n+16)=pi^2/a-b where a in N and b in Q , then a is equal to Given that sum(n=1)^oo 1/n^2=pi^2/6 and sum(n=1)^oo 1/(n^2+8n+16)=pi^2/a-b where a in N and b in Q , then a is equal to](https://d10lpgp6xz60nq.cloudfront.net/ss/web/60853.jpg)
Given that sum(n=1)^oo 1/n^2=pi^2/6 and sum(n=1)^oo 1/(n^2+8n+16)=pi^2/a-b where a in N and b in Q , then a is equal to
![Given that `sum_(n=1)^oo 1/n^2=pi^2/6 and sum_(n=1)^oo 1/(n^2+8n+16)=pi^2/a-b` where a in `N ... - YouTube Given that `sum_(n=1)^oo 1/n^2=pi^2/6 and sum_(n=1)^oo 1/(n^2+8n+16)=pi^2/a-b` where a in `N ... - YouTube](https://i.ytimg.com/vi/GA_KdA904fM/maxresdefault.jpg)
Given that `sum_(n=1)^oo 1/n^2=pi^2/6 and sum_(n=1)^oo 1/(n^2+8n+16)=pi^2/a-b` where a in `N ... - YouTube
![sequences and series - Using roots of unity to prove that $\cos\frac{\pi }{2n}\cos\frac{2\pi}{2n}\cdots\cos\frac{(n-1)\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$ - Mathematics Stack Exchange sequences and series - Using roots of unity to prove that $\cos\frac{\pi }{2n}\cos\frac{2\pi}{2n}\cdots\cos\frac{(n-1)\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$ - Mathematics Stack Exchange](https://i.stack.imgur.com/plfM7.jpg)